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3 years ago

PLEASE HELP ASAP WILL GIVE BRAINLIES

Mathematics

1 answer:

Lisa [10]3 years ago

8 0

Answer:

17: 480

(We first arrange all of the candidates except b in a line. This can be done in 5! ways. We now add b to the line. Since we now want to count the arrangements where b is positioned next to a, there are two possibilities for the position of b. (Either left of a or right of a).

There are thus 2⋅5! ways of arranging the candidates so that a and b are next to each other, and hence 6!−2⋅5!=4⋅5!=480 ways of arranging the candidates so that a and b are not next to each other)

18: i don't know sorry

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Step-by-step explanation:

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3 years ago

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The sum of the diagonals of a rhombus is 5√2.

Alex

Greetings!
Let ABCD be a rhombus
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Squaring both sides, we get
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In △AOB, we have
$OA {}^{2} + OB {}^{2} =AB {}^{2}$
$( \frac{ AC }{2} ) {}^{2} + ( \frac{ BD}{2} ) {}^{2} = AB {}^{2}$
$\frac{ AC {}^{2} } {4} + \frac{ BD {}^{2} }{4} = AB {}^{2}$
$AC {}^{2} + BD {}^{2} = 4 AB {}^{2}$
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Square rooting both sides
$\sqrt{34} = 2 AB$
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3 years ago

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Answer:

Step-by-step explanation:

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